Is the infinite product for sin(x) correct?

[Werg22]

Is this true?

f(x)=sin(x)=x(x-2pi)(x-4pi)(x-6pi)...

edit:

f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

 

[TD]

It doesn't seem right to me, try x = 2pi.

 

[Werg22]

Sorry I did a mistake. I meant sin(x).

 

[TD]

Try x = pi.

 

[Werg22]

I'm affraid I do not see what you mean... the serie is infinite...

 

[shmoe]

No, your infintite product does not even converge.

Infinite product form for sine:

$$\sin(x)=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{(\pi n)^2} \right)$$

 

[TD]

I'm affraid I do not see what you mean... the serie is infinite...

I was trying to point out that sin(pi) = 0 but your serie would never go to 0.

 

[Werg22]

Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.

 

[shmoe]

Sorry what I really meant is f(x)=sin(x)=x(x-pi)(x-2pi)(x-3pi)(x-4pi)...

Really sorry.

Still doesn't converge, the absolute value of the terms is growing without bound.